Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: functions, domains and operations (201-SN2-RE)

This sheet does not go over the course on functions again: you already have it, and it often dates back to high school. It deals with what decides the mark later in the semester, when a domain set up wrong makes the limits, the asymptotes and the optimization answers that fall outside the admissible interval wrong.

It is written for CEGEP students in the Science program (Sciences de la nature) who are starting Differential Calculus, 201-SN2-RE (formerly 201-NYA). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The domain is not a formality at the top of your solution: it decides whether a limit makes sense, whether a composition exists, and whether the answer to a problem is acceptable. Every mark lost in this chapter comes from a forgotten condition, never from a botched calculation.

This chapter is part of 201-SN2-RE, Differential Calculus

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Domain of a function
Range of a function
Absolute value
Piecewise-defined function
Composition of functions
Operations on functions
Transformations of graphs
Inverse function
One-to-one function
Unit circle
Reference angle
Physical domain of a model

The essentials

Three restrictions, and only three

• A DENOMINATOR is never zero.
• A RADICAND with an even index is greater than or equal to zero.
• The ARGUMENT of a logarithm is strictly positive.
• Everything else in $\mathbb{R}$ is allowed. A numerator that is zero, a square, a sine restrict nothing.
• When a root moves to the DENOMINATOR, the first two restrictions add up and the inequality becomes STRICT.

Same expression

9-x^{2}

, two domains:

\sqrt{9-x^{2}}

reaches

0

\pm 3

and includes them, while

\frac{1}{\sqrt{9-x^{2}}}

blows up at these two points and excludes them.

Writing the conditions BEFORE solving them, one per line, earns the method mark even if the final intersection goes wrong.

The domain of each operation, line by line

• $\text{dom}(f+g)=\text{dom}(f\cdot g)=\text{dom}\,f\cap\text{dom}\,g$ : both functions must exist.
• $\text{dom}\left(\frac{f}{g}\right)=\left(\text{dom}\,f\cap\text{dom}\,g\right)\setminus\left\{x:g(x)=0\right\}$ : remove the zeros of the DENOMINATOR.
• $\text{dom}(f\circ g)=\left\{x\in\text{dom}\,g\;:\;g(x)\in\text{dom}\,f\right\}$ : two conditions, in that order.
• $\text{dom}\,f^{-1}=\text{range}\,f$ and $\text{range}\,f^{-1}=\text{dom}\,f$ : the two sets swap.

The domain of a composition is NEVER read off the simplified expression: it is built from the two original functions.

Transformations, and the direction of each

• In $y=a\,f\left(b(x-c)\right)+d$ : what touches $x$ acts the opposite way, what is outside acts the natural way.
• $c>0$ shifts to the RIGHT; $b>1$ COMPRESSES horizontally; $a>1$ stretches vertically; $a<0$ reflects in the $x$ -axis.
• A point $(p, q)$ on the graph of $f$ becomes $\left(\frac{p}{b}+c, \;aq+d\right)$ .
• A NEGATIVE coefficient $a$ reverses the range interval: $\left[0, 5\right]$ becomes $\left[-7, 3\right]$ under $q\mapsto-2q+3$ .

With

a=2

b=1

c=1

d=1

, the point

(4, 2)

y=\sqrt{x}

becomes

(4+1, 2\cdot 2+1)=(5, 5)

: the shift inside the parentheses acts on

x

, the outside factor and term act on

y

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Forgetting that a root in the denominator requires a strict inequality

1 mark, and the vertical asymptotes missed afterwards

What not to write

“ $\text{dom}\left(\frac{1}{\sqrt{9-x^{2}}}\right)=\left[-3, 3\right]$ .”

What to write

“ $9-x^{2}>0$ , so $\text{dom}=\left(-3, 3\right)$ .”

Why: At $\pm 3$ the radicand is $0$ , an acceptable value under a root but forbidden in a denominator. These two points are exactly where the function goes to infinity: excluding them from the domain already announces two vertical asymptotes.

2. Believing that the domain of a composition is that of the inner function

2 marks, and a vertical asymptote that goes undetected

What not to write

“ $g(x)=\sqrt{x}$ has domain $\left[0, +\infty\right)$ , so $f\circ g$ does too.”

What to write

“We also need $g(x)\in\text{dom}\,f$ : here $\sqrt{x}\neq 2$ , so $x\neq 4$ , and $\text{dom}(f\circ g)=\left[0, 4\right)\cup\left(4, +\infty\right)$ .”

\frac{1}{\sqrt{x}-2}

exists for every

x\ge 0

EXCEPT at

x=4

, where the curve goes to infinity on both sides: this exclusion comes from the outer function, not from the root. It is an asymptote, not a hole.

Why: The two-step method is not a write-up precaution: it finds an excluded point that nothing in the domain of $g$ announced. It is the only safe way not to miss it.

3. Writing that the square root of a square equals the variable

2 marks here, and half a question on limits later

What not to write

$\sqrt{x^{2}}=x$

What to write

$\sqrt{x^{2}}=|x|$ , so $x$ if $x\ge 0$ and $-x$ if $x<0$

y=\sqrt{x^{2}}

is the V of

y=|x|

(green): to the left of

0

, the curve goes back up. The line

y=x

(red) follows the right path on the right side only.

Why: $\sqrt{(-2)^{2}}=\sqrt{4}=2$ , not $-2$ . This identity is what produces TWO horizontal asymptotes for $\frac{\sqrt{x^{2}+1}}{x}$ , and forgetting it is the costliest error of the next chapter.

4. Removing the zeros of the numerator instead of those of the denominator

1 mark, plus the real forbidden values that remain unreported

What not to write

“ $f(x)=\frac{x+3}{x^{2}-4}$ is not defined at $x=-3$ .”

What to write

“ $f$ is not defined at $x=\pm 2$ , where the denominator is zero; at $x=-3$ it equals $0$ .”

x=-3

, the NUMERATOR is zero: the curve crosses the axis, it is a zero of

f

. The forbidden values are

x=\pm 2

, where the denominator is zero and where the asymptotes start.

Why: A zero numerator gives a perfectly legitimate value, a ZERO of the function. Confusing a zero with a forbidden value then leads you to look for an asymptote where the curve simply crosses the axis.

5. Shifting in the wrong direction

1 mark, and the whole reading of the graph that follows

What not to write

“ $f(x-1)$ shifts the curve $1$ unit to the left.”

What to write

“ $f(x-1)$ shifts the curve $1$ unit to the RIGHT: the new curve reaches at $x=1$ what $f$ reached at $0$ .”

f(x)=x^{2}

has its vertex at

0

;

f(x-1)

has it where the parentheses are zero, at

x=1

: to the RIGHT (green). The shift to the left (red) is that of

f(x+1)

Why: The five-second test that settles it: which $x$ -value makes the parentheses zero? For $f(x-1)$ , it is $x=1$ , so the point that was at $0$ ends up at $1$ , that is, to the right. This test beats a memorized rule, which flips one time out of two under exam pressure.

6. Keeping the order of the endpoints when the coefficient is negative

1 mark, and an interval that means nothing

What not to write

“ $\text{range}\,f=\left[0, 5\right]$ , so $\text{range}\left(-2f+3\right)=\left[3, -7\right]$ .”

What to write

“The transformation $q\mapsto-2q+3$ is DECREASING, it swaps the endpoints: $\text{range}=\left[-7, 3\right]$ .”

The line

q\mapsto -2q+3

goes DOWN: the endpoint

0

[0, 5]

(blue) becomes the UPPER endpoint

3

, and

5

becomes the lower one,

-7

. The range is read on the vertical axis, bottom to top:

[-7, 3]

Why: An interval is always written from the smallest to the largest. The real content of the question lies elsewhere: a decreasing transformation sends the lower endpoint to the upper endpoint, and you have to say so, not just reorder the numbers at the end.

7. Believing that every function has an inverse

the whole question, often 3 marks

What not to write

“ $g(x)=x^{2}-4x$ , I isolate $x$ , so $g^{-1}$ exists.”

What to write

“ $g(0)=g(4)=0$ : $g$ is not one-to-one on $\mathbb{R}$ . On $\left[2, +\infty\right)$ , $g^{-1}(x)=2+\sqrt{x+4}$ ; on $\left(-\infty, 2\right]$ , $g^{-1}(x)=2-\sqrt{x+4}$ .”

The parabola

x^{2}-4x

takes the value

0

twice (at

0

and at

4

): no inverse on

\mathbb{R}

. On the green branch,

x\ge 2

, each height is reached only once.

Why: Isolating $x$ produces a $\pm$ in front of the root, so two values, so no function. The sign is chosen from the RESTRICTION adopted, never out of habit: it is the same reasoning that forces $\arcsin$ to be defined on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

8. Taking the algebraic expression for the function of the problem

the whole optimization question if the answer falls outside the domain

What not to write

“ $V(x)=x(30-2x)(40-2x)$ is a polynomial, so its domain is $\mathbb{R}$ .”

What to write

“The physically admissible domain is $\left(0, 15\right)$ , because we need $x>0$ , $30-2x>0$ and $40-2x>0$ .”

The formula

V(x)=x(30-2x)(40-2x)

can be computed everywhere, but only the green part,

0<x<15

, is a box. Beyond it, it gives negative volumes, such as

V(16)=-256

Why: $V(16)=-256$ : the formula can still be computed and gives a negative volume. The physics is carried by the DOMAIN, not by the formula, and the most restrictive constraint wins over the others.

9. Removing values from a denominator that is never zero

1 mark, and a vertical asymptote invented in the next chapter

What not to write

“ $f(x)=\frac{x-4}{3x^{2}-4x+7}$ is a fraction, so the roots of the denominator must be excluded.”

What to write

“The discriminant of $3x^{2}-4x+7$ is $16-84=-68<0$ : the quadratic is never zero and keeps the sign of its leading coefficient, positive. So $\text{dom}\,f=\mathbb{R}$ .”

3x^{2}-4x+7

stays above the axis (discriminant

-68<0

, minimum

\frac{17}{3}

): the denominator is never zero, there is no value to exclude.

Why: The presence of a fraction is not in itself a restriction: the restriction comes from the denominator ACTUALLY being zero, which must therefore be checked instead of assumed. The reflex of writing “forbidden values” before computing the discriminant produces imaginary exclusions, and later a vertical asymptote that does not exist. Three quotients out of four do have forbidden values; it is the fourth one that the exam puts on the list.

10. Declaring two functions equal because their rules simplify into each other

1 mark here, the nature of a discontinuity later

What not to write

“ $p(x)=\frac{x^{2}-4}{x-2}$ and $q(x)=x+2$ are the same function.”

What to write

“Their rules agree for $x\neq 2$ , but $\text{dom}\,p=\mathbb{R}\setminus\{2\}$ and $\text{dom}\,q=\mathbb{R}$ : they are not the same functions.”

Why: Two functions are equal if and only if they have the SAME domain AND the same rule. This is not vocabulary: the graph of $p$ is that of $q$ with a hole at $(2, 4)$ , and simplifying hid the restriction without removing it. That is exactly what separates a hole from a vertical asymptote when a factor of the denominator cancels with the numerator, and it is why the domain is always read on the ORIGINAL expression.

Which method to choose

How to build a domain, condition by condition

Scan the expression and write down ONE condition per risky element, without solving anything. Solving comes next, and the intersection last.

If there is a quotient → set denominator $\neq 0$ and solve

Example: $\frac{x+3}{x^{2}-4}$ gives $x\neq\pm 2$
If there is a root with an even index, outside the denominator → set radicand $\ge 0$ , NON-STRICT inequality

Example: $\sqrt{6-2x}$ gives $x\le 3$
If there is a root with an even index IN the denominator → set radicand $>0$ , STRICT inequality

Example: $\frac{1}{\sqrt{9-x^{2}}}$ gives $-3<x<3$
If there is a logarithm → set argument $>0$

Example: $\ln(x-1)$ gives $x>1$
If there is a composition $f(g(x))$ → two conditions: $x\in\text{dom}\,g$ , then $g(x)\in\text{dom}\,f$

Example: $\frac{1}{\sqrt{x}-2}$ gives $x\ge 0$ AND $x\neq 4$
If it is a real-world problem → add all the physical constraints and keep the most restrictive one

Example: cutting the corners gives $x>0$ , $x<15$ and $x<20$ , so $\left(0, 15\right)$

A quadratic under a root is handled with a sign chart: positive OUTSIDE its roots when the leading coefficient is positive. That is where the left half of the domain gets lost.

Which method for a trigonometry question

The form of the question points to the tool, and there are only three.

\frac{4\pi}{3}

falls in the third quadrant, symmetric to

\frac{\pi}{3}

through the origin: same reference angle, but the

x

-coordinate of the point is negative,

\cos\frac{4\pi}{3}=-\frac{1}{2}

If you are asked for an exact value at a non-standard angle → reference angle, then the sign given by the quadrant

Example: $\cos\frac{4\pi}{3}$ : reference $\frac{\pi}{3}$ , third quadrant, cosine negative, so $-\frac{1}{2}$
If you are asked to solve an equation over one turn → reference angle, then the TWO quadrants where the function has the right sign

Example: $\cos x=\frac{\sqrt{3}}{2}$ gives $\frac{\pi}{6}$ and $\frac{11\pi}{6}$
If you are asked for the amplitude, period or phase shift → FACTOR OUT the coefficient of $x$ first, then read on $a\sin\left(b(x-c)\right)+d$

Example: $3\sin\left(2x-\frac{\pi}{3}\right)+1=3\sin\left(2\left(x-\frac{\pi}{6}\right)\right)+1$ , phase shift $\frac{\pi}{6}$

A trigonometric equation has two families of solutions per turn, except at the extreme values $\pm 1$ . Giving only one costs half the question.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a domain the way the marker expects

When to use it: Every time you write a domain, including when the question does not explicitly ask for it.

1 List the conditions, one per line, before solving anything: “We need $x^{2}-x-6\ge 0$ and $x-4\neq 0$ .”
2 Solve each condition separately, showing the sign chart for a quadratic.
3 Combine the conditions and write the intersection in INTERVAL notation, with the right brackets.
4 Check with two test values: one that must be inside, one that must be outside.

Concluding sentence

$\text{dom}\,h=\left(-\infty, -2\right]\cup\left[3, 4\right)\cup\left(4, +\infty\right)$ .

The trap: Writing the domain in words (“all real numbers except 4 and the negative values”) instead of a union of intervals. The notation is what is marked, and a wrong bracket, a parenthesis where a square bracket belongs or the reverse, costs as much as a forgotten condition.

Marking: Usually half a mark per condition set up, half a mark per condition solved, one mark for the intersection correctly written.

Writing up the search for an inverse

When to use it: The question asks for $f^{-1}$ , or to “express $x$ in terms of $y$ ”.

1 First justify that the function is one-to-one, or restrict it: “ $f$ is strictly increasing on $\left[2, +\infty\right)$ , so it is one-to-one.”
2 Set $y=f(x)$ and isolate $x$ , keeping track of the sign conditions.
3 Choose the sign of the root from the restriction adopted, and SAY so.
4 Swap the names of the variables to write $f^{-1}(x)$ .
5 Give the domain and the range of $f^{-1}$ , recalling that they are those of $f$ swapped.

Concluding sentence

On $\left[2, +\infty\right)$ , $g^{-1}(x)=2+\sqrt{x+4}$ , with domain $\left[-4, +\infty\right)$ and range $\left[2, +\infty\right)$ .

The trap: Skipping the one-to-one justification. An inverse written without it is wrong as soon as the function is not monotonic, and the marker takes the mark off even if the formula is right.

Marking: The one-to-one argument or the restriction is almost always worth one mark on its own.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Test two values, one inside and one outside

Choose a number that should belong to the domain and check that the calculator returns a value, then a number that should be excluded and check that it returns an error.

On $\frac{\sqrt{x^{2}-x-6}}{x-4}$ : $x=5$ gives $\sqrt{14}$ , so $5$ is inside ✓; $x=0$ requires $\sqrt{-6}$ , so $0$ is outside ✓.
Check an inverse by composing

Compute $f\left(f^{-1}(c)\right)$ for any $c$ in the domain: the result must be exactly $c$ .

$f^{-1}(5)=-\frac{16}{3}$ and $f\left(-\frac{16}{3}\right)=5$ ✓. A composition that does not give back $c$ signals an error in isolating $x$ , often a sign.
Check a transformation on a single point

Take a key point of the original graph, apply the rule $(p, q)\mapsto\left(\frac{p}{b}+c, \;aq+d\right)$ to it, and check it on the sketch.

The vertex $(0, 0)$ of $y=x^{2}$ becomes $(1, -4)$ for $y=(x-1)^{2}-4$ : the sketch must show a vertex at $(1, -4)$ , not at $(-1, 4)$ .
Reread the unit and the meaning in a real-world problem

Substitute a value from the domain and check that the answer makes physical sense: a positive length, a positive volume, a quantity smaller than the stock.

$V(16)=-256$ : a negative volume proves that $16$ does not belong to the admissible domain, even though the polynomial can be computed.

The typical problem, taken apart

A domain with two restrictions, as on the first test

Find the domain of $h(x)=\dfrac{\sqrt{x^{2}-x-6}}{x-4}$ , in interval notation.

The parabola

x^{2}-x-6

is positive OUTSIDE its roots

-2

and

3

: the left half of the domain is the one students forget. The vertical line at

x=4

marks the second restriction, independent of the first.

Step 1

List the conditions before solving: the root requires $x^{2}-x-6\ge 0$ , and the denominator requires $x-4\neq 0$ .

Why

Writing the two conditions separately keeps you from forgetting one, and earns the method mark even if the rest goes off track. The two restrictions are independent: neither follows from the other.

Step 2

Factor the quadratic: $x^{2}-x-6=(x-3)(x+2)$ , whose roots are $-2$ and $3$ . Since the leading coefficient is positive, the quadratic is positive OUTSIDE its roots: $x\le-2$ or $x\ge 3$ .

Why

This is where the left half of the domain gets lost. The sentence “positive outside the roots” should be written as is: it is worth the justification mark and stops you from keeping only $\left[3, +\infty\right)$ .

Step 3

The inequalities are NON-STRICT, because $\sqrt{0}=0$ exists: the values $-2$ and $3$ belong to the domain.

Why

The root is in the numerator, so nothing forbids it from being zero. Had it been in the denominator, the same two values would have been excluded and the square brackets would have become parentheses.

Step 4

Second condition: $x\neq 4$ . Now $4$ belongs to $\left[3, +\infty\right)$ , so it must indeed be removed.

Why

A condition only has an effect if it bears on a point already kept. Checking this avoids writing a useless exclusion, which betrays a method applied mechanically.

Step 5

Intersection: $\text{dom}\,h=\left(-\infty, -2\right]\cup\left[3, 4\right)\cup\left(4, +\infty\right)$ .

Why

Three intervals, three different pairs of brackets. Each bracket tells a condition: a square bracket for a root that can be zero, a parenthesis for a denominator that is zero.

Step 6

Check: $h(5)=\frac{\sqrt{14}}{1}\approx 3.74$ exists ✓; $h(0)$ requires $\sqrt{-6}$ , so $0$ is indeed excluded ✓; $h(-3)=\frac{\sqrt{6}}{-7}$ exists ✓, which confirms the left half.

Why

The test at $x=-3$ is the most useful of the three: it checks precisely the part of the domain that the classic error deletes.

The conclusion, written out

$\text{dom}\,h=\left(-\infty, -2\right]\cup\left[3, 4\right)\cup\left(4, +\infty\right)$ , the square brackets coming from the root, which can be zero, and the parentheses around 4 from the denominator, which cannot.

The classic mistake on this problem: Keeping only $\left[3, 4\right)\cup\left(4, +\infty\right)$ , forgetting that the quadratic is also positive for $x\le-2$ . Half of the domain disappears, and with it every later question on that branch of the curve.

Learn by heart

• Three restrictions: nonzero denominator, nonnegative radicand for an even root, strictly positive argument of a logarithm. Nothing else.
• Root IN the denominator: the two add up and the inequality becomes strict.
• $\sqrt{x^{2}}=|x|$ , never $x$ . This identity will produce two horizontal asymptotes in the next chapter.
• $\text{dom}(f\circ g)$ : $x\in\text{dom}\,g$ AND $g(x)\in\text{dom}\,f$ . Two conditions, never one.
• A quadratic with a positive leading coefficient is positive OUTSIDE its roots.
• $f(x-c)$ shifts to the right when $c>0$ : the $x$ -value that makes the parentheses zero tells you everything.
• A transformation with a negative coefficient REVERSES the range interval.
• No inverse unless the function is one-to-one. Otherwise restrict it, and the restriction chooses the sign of the root.
• Two functions are equal if and only if they have the same domain AND the same rule. Simplifying an expression never changes the domain.

Frequently asked questions

How do you find the domain of a function?

Look for the only three restrictions: a denominator that is zero, a negative radicand under an even root, a logarithm whose argument is negative or zero. Set one condition per risky element, solve each one separately, then take the intersection and write it in interval notation.

Why does the domain change when the root is in the denominator?

Because two restrictions add up. Under a root alone, the radicand can be zero, and the endpoints belong to the domain. In the denominator, the root must not be zero either, so the radicand must be strictly positive and the endpoints leave the domain.

Is the domain of a composition the domain of the inner function?

No, it is usually smaller. The variable must belong to the domain of the inner function, and in addition the output of that function must belong to the domain of the outer function. This second condition often brings out an excluded point that nothing announced.

In which direction does f of x minus 1 shift the graph?

To the right. The five-second test is to find the x-value that makes the parentheses zero: here x equals one, so what the function did at zero now happens at one, that is, further right. What touches the variable always acts against intuition.

When does a function have an inverse?

When it is one-to-one, that is, when each output comes from only one input. A parabola is not, since a horizontal line cuts it twice. It is then restricted to an interval where it is strictly monotonic, and that restriction decides the sign in front of the root in the inverse.

Why do you need a physical domain in an optimization problem?

Because the algebraic formula can still be computed outside the physical meaning. The volume of a box cut from a sheet of cardboard gives a negative number for a cut that is too large. The physics is carried by the domain, not by the formula, and an answer that falls outside the admissible domain is worth nothing.

Practise it

Corrected exercises: Functions, domains and operations, 201-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

17 corrected exercises
170 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Next sheet Limits and continuity

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc1-functions-and-domains. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: functions, domains and operations (201-SN2-RE)

The essentials

Three restrictions, and only three

The domain of each operation, line by line

Transformations, and the direction of each

The mistakes that cost marks

1. Forgetting that a root in the denominator requires a strict inequality

2. Believing that the domain of a composition is that of the inner function

3. Writing that the square root of a square equals the variable

4. Removing the zeros of the numerator instead of those of the denominator

5. Shifting in the wrong direction

6. Keeping the order of the endpoints when the coefficient is negative

7. Believing that every function has an inverse

8. Taking the algebraic expression for the function of the problem

9. Removing values from a denominator that is never zero

10. Declaring two functions equal because their rules simplify into each other

Which method to choose

How to build a domain, condition by condition

Which method for a trigonometry question

How the answer is expected to be written

Writing up a domain the way the marker expects

Writing up the search for an inverse

Check before you hand in

The typical problem, taken apart

A domain with two restrictions, as on the first test

Learn by heart

Frequently asked questions

How do you find the domain of a function?

Why does the domain change when the root is in the denominator?

Is the domain of a composition the domain of the inner function?

In which direction does f of x minus 1 shift the graph?

When does a function have an inverse?

Why do you need a physical domain in an optimization problem?

Corrected exercises: Functions, domains and operations, 201-SN2-RE

See also

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